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Equation Icon  ( 2Δ2 1 ) Pab = 2BacBbd γcd + ---- − --B2 γab, (113 ) 3 3 4Δ2 B2 E = ---- + ---, (114 ) 3 3 d ⋆3 B = − ⋆3 ihB − ⋆3(dΔ − Δh ) + 4√ξ-ΔB, (115 ) 3